So let’s take this and consider a linear transformation . The first isomorphism theorem says we can factor as a surjection followed by an injection . We’ll just regard the latter as the inclusion of the image of as a subspace of . As for the surjection, it must be the linear map , just as in any abelian category. Then we can set up the short exact sequence

So where do “rank” and “nullity” come in? Well, these are just jargon terms. The “rank” of a linear transformation is the dimension of its image — not the target vector space, mind you, but the subspace of vectors of the form . That is, it’s the size of the largest set of linearly independent vectors in the image of the transformation. The “nullity” is the dimension of the kernel — the largest number of linearly independent vectors that sends to the zero vector in .

So what does the direct sum decomposition above mean? It tells us that there is a basis of which is in bijection with the disjoint union of a basis for and a basis for . In the finite-dimensional case we can take cardinalities and say that the dimension of is the sum of the dimensions of and . Or, to use our new words, the dimension of is the sum of the rank and the nullity of . Thus: the rank-nullity theorem.

[…] be the dimension of this subspace, which we called the nullity of the linear transformation . The rank-nullity theorem then tells us that we have a relationship between the number of independent solutions to the system […]

[…] system has no solutions at all. What’s the problem? Well, we’ve got a linear map . The rank-nullity theorem tells us that the dimension of the image (the rank) plus the dimension of the kernel (the nullity) […]

[…] the dimension of the cokernel add up to the dimension of the target space. But notice also that the rank-nullity theorem tells us that the dimension of the kernel and the dimension of the image add up to the dimension of […]

[…] count the dimensions — if has dimension then each space has dimension — and use the rank-nullity theorem to see that they must be isomorphic. That is, every -multilinear functional is a linear combination […]

[…] that this is onto, so the dimension of the image is . The dimension of the source is , and so the rank-nullity theorem tells us that the dimension of the kernel — the dimension of the space that sends back to […]

[…] to a constant function is automatically zero. Thus we conclude that . In fact, we can say more. The rank-nullity theorem tells us that the dimension of and the dimension of add up to the dimension of , which of course […]

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This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.